Challenge 9

Interactive Preview of “StateNames_A” Dataset

Code

datatable(StateNames_A)

3 Summarizing & Visualizing the Number of Allisons

Make a summary table of the number of babies named “Allison” for each state and the sex of the baby. Specifically, the table should have the following qualities:

each state should be its own row
and each sex should have its own column
if there were no babies born for that combination of state & sex there should be a 0 (not an NA)

Code

Allison_table <- StateNames_A |> 
  filter(Name == 'Allison') |>
  rename(sex = Gender) |>
  group_by(sex, State) |>
  summarise(total_count = sum(Count)) |>
  pivot_wider(names_from = sex,
              values_from = total_count) |>
  replace_na(list('M' = 0,
             'F' = 0))

Allison_table |> 
  kable(format = "html", 
        col.names = 
          c("State", 
            "Females", 
            "Males"),
        caption = "Number of Babies named 'Allison' in each State sorted by Sex")

Number of Babies named 'Allison' in each State sorted by Sex
State	Females	Males
AK	232	0
AL	1535	0
AR	1198	0
AZ	1880	0
CA	12413	0
CO	1594	0
CT	1099	0
DC	321	0
DE	294	0
FL	4455	0
GA	3257	0
HI	183	0
IA	1477	0
ID	451	0
IL	5110	0
IN	3067	0
KS	1283	0
KY	1905	20
LA	1209	0
MA	2218	0
MD	2229	0
ME	340	0
MI	4014	0
MN	2374	0
MO	2882	0
MS	817	0
MT	226	0
NC	3435	0
ND	285	0
NE	807	0
NH	412	0
NJ	3052	0
NM	399	0
NV	729	0
NY	5747	0
OH	5487	0
OK	1421	0
OR	1186	0
PA	4307	0
RI	306	0
SC	1228	0
SD	376	0
TN	2488	0
TX	10192	0
UT	1125	0
VA	3220	0
VT	135	0
WA	1956	0
WI	2367	0
WV	813	0
WY	142	0

You should have seen in the above that “Allison” is a name given overwhelmingly to babies assigned “female” at birth. So, create a new dataset named allison_f which contains only the babies assigned Female at birth.

Code

allison_f <- StateNames_A |> 
  rename(sex = Gender) |>
  filter(Name == 'Allison', sex == 'F') |>
  group_by(State) |>
  replace_na(list('Count' = 0)) |>
  select(Year, State, Count)

allison_f |> 
  kable(format = "html",
        col.names = 
          c("Name", 
            "Year", 
            "Sex",
            "State",
            "Number of Babies"),
        caption = "Number of Female Babies named 'Allison' per State by Year")

Error in dimnames(x) <- dn: length of 'dimnames' [2] not equal to array extent

Make a visualization showing how the popularity of the name “Allison” has changed over the years. To be clear, each year should have one observation–the total number of Allisons born that year.

Code

ggplot(data = allison_f, mapping = aes(x = Year, y = Count)) +
  geom_jitter(alpha = 0.3, color = 'tomato') +
  labs(title = "Number of Allisons in State over Time")

4 Modeling the Number of Allisons

Fit a linear model with the year as the explanatory variable, and the number of Allisons as the response. Similar to #3, each year should have one observation–the total number of Allisons born that year.
Code
```
allison_lm <- lm(Count ~ Year, data = allison_f)
```

Write out the estimated regression equation.

Code

tidy(allison_lm)

# A tibble: 2 × 5
  term        estimate std.error statistic p.value
  <chr>          <dbl>     <dbl>     <dbl>   <dbl>
1 (Intercept)  3894.    1766.         2.21  0.0277
2 Year           -1.88     0.880     -2.14  0.0328

Estimated Number of Allisons = 3892.55986 - 1.881963(Year)

Plot the residuals of the model, that is, the actual values minus the predicted values. Comment on the residuals - do you see any patterns?
Code
```
allison_lm |> 
  augment() |> 
  ggplot(mapping = aes(y = .resid, x = .fitted)) +
  geom_point()
```
It appears that that there is not equal variance present in this model because there are clearly far more outlying positive residuals lying above the least squares line. Additionally, a pattern I can see is that the variance is decreasing as the fitted values increase, as the residual points tend to come closer to 0.
What do you conclude from this model? Is my name not cool anymore?

It is difficult to make conclusions from this model since equal variance is not present.

5 Spelling by State

Narrow the A name dataset (downloaded previously) down to only male-assigned babies named “Allan”, “Alan”, or “Allen”. Make a plot comparing the popularity of these names over time.

Code

Aln_data <- StateNames_A |> 
  filter(Name == 'Allan' | 
         Name == 'Alan' | 
         Name == 'Allen') |>
  rename(sex = Gender) |>
  group_by(Name, Year) |>
  summarise(total_count = sum(Count))

Aln_data |> 
  kable(format = "html",
        col.names = 
          c("Name", 
            "Year", 
            "Number of Babies"),
        caption = "Number of Babies per Spelling of 'Allen' by Year")

Number of Babies per Spelling of 'Allen' by Year
Name	Year	Number of Babies
Alan	1997	2155
Alan	1998	2102
Alan	1999	2220
Alan	2000	2398
Alan	2001	2622
Alan	2002	2591
Alan	2003	3083
Alan	2004	3051
Alan	2005	3189
Alan	2006	3442
Alan	2007	3225
Alan	2008	3002
Alan	2009	2817
Alan	2010	2487
Alan	2011	2321
Alan	2012	2252
Alan	2013	2577
Alan	2014	2464
Allan	1997	430
Allan	1998	455
Allan	1999	457
Allan	2000	435
Allan	2001	441
Allan	2002	443
Allan	2003	576
Allan	2004	561
Allan	2005	509
Allan	2006	473
Allan	2007	466
Allan	2008	436
Allan	2009	436
Allan	2010	363
Allan	2011	299
Allan	2012	287
Allan	2013	328
Allan	2014	282
Allen	1997	1371
Allen	1998	1289
Allen	1999	1296
Allen	2000	1319
Allen	2001	1364
Allen	2002	1365
Allen	2003	1286
Allen	2004	1273
Allen	2005	1199
Allen	2006	1201
Allen	2007	1163
Allen	2008	1120
Allen	2009	1004
Allen	2010	935
Allen	2011	972
Allen	2012	924
Allen	2013	873
Allen	2014	828

Code

ggplot(data = Aln_data, mapping = aes(x = Year, y = total_count)) +
  geom_line() +
  facet_wrap(~ Name) +
  labs(title = "Total Number of Allan, Alan, and Allens over Time")

In California, Allan’s spelling of his name is the least common of the three but perhaps it’s not such an unusual name for his home state of Pennsylvania. Compute the total number of babies born with each spelling of “Allan” in the year 2000, in Pennsylvania and in California. Specifically, the table should have the following qualities:

each spelling should be its own column
each state should have its own row
a 0 (not an NA) should be used to represent locations where there were no instances of these names

Convert your total counts to overall percents. That is, what was the percent breakdown between the three spellings in CA? What about in PA?

Code

Aln_prop <- StateNames_A |>
  filter(Name %in% c('Allan', 'Alan', 'Allen'),
           Year == 2000,
           State %in% c('CA', 'PA')
         ) |>
  group_by(Name, State) |>
  summarize(total_count = sum(Count)) |>
  group_by(State) |>
  mutate(Prop = total_count / sum(total_count)) |>
  select(Name, State, Prop) |>
  pivot_wider(names_from = Name, values_from = Prop)

Aln_prop |> 
  kable(format = "html",
        digits = 3,
        col.names = 
          c("State", 
            "Alan", 
            "Allan",
            "Allen"),
        caption = "Proportions of Different Spellings of 'Allen' per State") |>
  kable_classic(html_font = "Ariel",
                font_size = 18) |>
  row_spec(0, bold = TRUE)

Proportions of Different Spellings of 'Allen' per State
State	Alan	Allan	Allen
CA	0.655	0.147	0.198
PA	0.429	0.101	0.471